1.2. Lecture 2

1.2.1. More results about the exponential map

For the next result about the exponential map it will be useful to know the following fact from linear algebra.

Lemma 1.2.1.

Let $X\in\operatorname{GL}_{n}(\mathbb{C})$ . Then $X$ is conjugate to a matrix of the form $DU$ where $D$ is diagonal, $U$ is upper triangular with ‘1’s on the diagonal and $D$ and $U$ commute.

Proof.

(nonexaminable) This follows from Jordan normal form. Here’s a direct proof. Firstly write $\mathbb{C}^{n}$ as a direct sum of generalised eigenspaces for $X$ : if $\lambda$ is an eigenvalue of $X$ then we can write the characteristic polynomial $P(T)=(T-\lambda)^{a}Q(T)$ where $Q(T)$ does not have $\lambda$ as a root and $a\geq 1$ is an integer. Then the image of $Q(X)$ on $\mathbb{C}^{n}$ is the generalised eigenspace of $\lambda$ . The kernel of $Q(X)$ is preserved by $X$ and $X$ does not have an eigenvalue equal to $\lambda$ since $\lambda$ is not a root of $Q(X)$ , which must be the characteristic polynomial of $X$ acting on $\ker Q(X)$ . Thus

\operatorname{im}Q(X)\cap\ker Q(X)=\{0\}

and by the rank-nullity theorem

\mathbb{C}^{n}=\operatorname{im}Q(X)\oplus\ker Q(X)

is a decomposition of $\mathbb{C}^{n}$ as a direct sum of the $\lambda$ generalised eigenspace and a subspace preserved by $X$ . Repeating for each eigenvector gives the required decomposition of $\mathbb{C}^{n}$ . This reduces the proof of the statement to the case where $X$ has only one eigenvalue $\lambda$ . In this case, we can inductively choose a basis ${\bf v}_{1},\ldots,{\bf v}_{n}$ of $\mathbb{C}^{n}$ such that, for $1\leq i\leq n$ , the image of ${\bf v}_{i}$ in $\mathbb{C}/\left\langle{\bf v}_{1},\ldots,{\bf v}_{i-1}\right\rangle$ is an eigenvector of $X$ with eigenvalue $\lambda$ . With respect to this basis, $X$ is then diagonal with $\lambda$ ’s on the diagonal, and we get the required decomposition with $D=\lambda\operatorname{Id}$ . ∎

Lemma 1.2.2.

The exponential function $\exp:\mathfrak{gl}_{n,\mathbb{C}}\to\operatorname{GL}_{n}(\mathbb{C})$ is surjective.

Proof.

First prove it for all $D$ and $U$ in $\operatorname{GL}_{n}(\mathbb{C})$ as in Lemma 1.2.1. The case of diagonal matrices is more straightforward (Problem 1.2.3(a)) whereas for $U$ you can use that the power series for $\log(U)$ in terms of powers of $U-\operatorname{Id}$ is actually a polynomial (Problem 1.2.3(b)).

Now consider $DU$ . If $D=\exp(d)$ and $U=\exp(u)$ then

DU=\exp(d)\exp(u)=\exp(d+u)

because $d$ and $u$ commute (so long as you choose $d$ and $u$ carefully — Problem 1.2.3(c)).

By Lemma 1.2.1 we have that for any $X$ in $\operatorname{GL}_{n}(\mathbb{C})$ there exists a $P$ in $\operatorname{GL}_{n}(\mathbb{C})$ such that $P^{-1}XP=DU$ where $D$ and $U$ have the form stated in the Lemma. By Lemma 1.1.2(v) we have

X=PDUP^{-1}=P\exp(d+u)P^{-1}=\exp(P(d+u)P^{-1})

and thus $\exp$ is surjective. ∎

Remark 1.2.3.

The lemma is not true over $\mathbb{R}$ ; as we will see, the determinant of $\exp(X)$ is positive for all real matrices $X$ .

Lemma 1.2.4.

We have

\det\exp(X)=\exp\operatorname{tr}(X).

Proof.

By Lemma 1.2.1 we can conjugate so that $X$ is an upper triangular matrix with diagonal entries $\lambda_{1},\ldots,\lambda_{n}$ , and then note that $\exp(X)$ is also upper triangular with diagonal entries $\exp(\lambda_{1}),\ldots,\exp(\lambda_{n})$ .

Thus

\det\exp(X)=\prod_{i=1}^{n}\exp(\lambda_{i})=\exp\left(\sum_{i=1}^{n}\lambda_{% i}\right)=\exp\operatorname{tr}(X).

∎

The next proposition will be useful when we discuss Lie algebras of linear Lie groups.

Proposition 1.2.5 (Lie product formula).

For $X,Y\in\mathfrak{gl}_{n,k}$ we have

\exp(X+Y)=\lim_{k\to\infty}\left(\exp\left(\frac{X}{k}\right)\exp\left(\frac{Y% }{k}\right)\right)^{k}.

Proof.

We have

	$\displaystyle\left(\exp\left(\frac{X}{k}\right)\exp\left(\frac{Y}{k}\right)% \right)^{k}$	$\displaystyle=\left(\left(\operatorname{Id}+\frac{X}{k}+O\left(\frac{1}{k^{2}}% \right)\right)\left(\operatorname{Id}+\frac{Y}{k}+O\left(\frac{1}{k^{2}}\right% )\right)\right)^{k}$
		$\displaystyle=\left(\operatorname{Id}+\frac{X+Y}{k}+O\left(\frac{1}{k^{2}}% \right)\right)^{k}$
		$\displaystyle=\operatorname{Id}+X+Y+O\left(\frac{1}{k}\right)$
		$\displaystyle=\exp\left(X+Y+O\left(\frac{1}{k}\right)\right).$

Now take the limit as $k\to\infty$ . ∎

1.2.2. One-parameter subgroups

Lemma 1.2.6.

The map from $\mathbb{R}$ to $\operatorname{GL}_{n}(\mathbb{C})$ given by

t\longmapsto\exp(tX)

is a differentiable group homomorphism.

We have

\frac{d}{dt}\exp(tX)=X\exp(tX)=\exp(tX)X.

In particular,

\left.\frac{d}{dt}\exp(tX)\right|_{t=0}=X.

Proof.

The given map is a group homomorphism by Lemma 1.1.2(iv).

By definition,

\exp(tX)=\sum_{k=0}^{\infty}\frac{X^{k}}{k!}t^{k}.

As this power series (and its termwise derivative) are uniformly convergent on any compact subset, we can compute its derivative by differentiating termwise, which gives

\frac{d}{dt}\exp(tX)=\sum_{k=1}^{\infty}\frac{X^{k}}{(k-1)!}t^{k-1}=X\exp(tX).\qed

Definition 1.2.7.

A one-parameter subgroup of $\operatorname{GL}_{n}(\mathbb{C})$ is a differentiable group homomorphism $f:\mathbb{R}\to\operatorname{GL}_{n}(\mathbb{C})$ . That is, a differentiable map such that

f(s+t)=f(s)f(t)

for all $s,t\in\mathbb{R}$ .

The infinitesimal generator of a one-parameter subgroup $f$ is the element $f^{\prime}(0)\in\mathfrak{gl}_{n,\mathbb{C}}$ .

The convention used here is a slight abuse of notation, $f(\mathbb{R})$ is the subgroup of $\operatorname{GL}_{n}(\mathbb{C})$ referred to in the definition but as the map defines the subgroup we just refer to the map.

Remark 1.2.8.

For a one-parameter subgroup $f$ , it actually suffices to require that $f$ is continuous. Differentiability then comes for free.

Indeed, if $f$ is continuous, the integral $\int_{0}^{a}f(t)dt$ exists. Moreover,

f(s)\int_{0}^{a}f(t)dt=\int_{0}^{a}f(s+t)dt=\int_{s}^{s+a}f(t)dt.

The RHS is differentiable with respect to $s$ by the fundamental theorem of algebra. Therefore, to prove that $f(s)$ is differentiable, we only need to show that there is an $a>0$ such that $\int_{0}^{a}f(t)dt$ is an invertible matrix. Now consider the function

F(a)=\frac{1}{a}\int_{0}^{a}f(t)dt.

It is well-defined for $a\neq 0$ and $\lim\limits_{a\to 0}F(a)=\operatorname{Id}$ . Hence, for $0<a\ll 1$ , $F(a)$ is invertible, and therefore so is $aF(a)=\int_{0}^{a}f(t)dt$ .

The following is a very important property of one-parameter subgroups: that they all come from the exponential map.

Proposition 1.2.9.

Let $f:\mathbb{R}\to\operatorname{GL}_{n}(\mathbb{C})$ be a one-parameter subgroup with infinitesimal generator $X$ .

Then

f(t)=\exp(tX)

for all $t\in\mathbb{R}$ . That is, all one-parameter subgroups arise from the exponential function.

Proof.

From the definition of one-parameter subgroups, we have

f^{\prime}(t)=\lim_{s\to 0}\frac{f(s+t)-f(t)}{s}=f(t)\lim_{s\to 0}\frac{f(s)-f% (0)}{s}=f(t)f^{\prime}(0)=f(t)X.

Now consider the differential equation

f^{\prime}(t)=f(t)X.

By Lemma 1.2.6 we have $\exp(tX)$ is a solution with the initial condition that $f(0)=\operatorname{Id}$ . To show it is a unique solution suppose that $g(t)$ is also a solution. Then

\left(g(t)\exp(-tX)\right)^{\prime}=g^{\prime}(t)\exp(-tX)-g(t)\exp(-tX)X=g(t)% \left(X\exp(-tX)-\exp(-tX)X\right)=0,

and thus $g(t)\exp(-tX)=D\in\operatorname{GL}_{n}(\mathbb{C})$ . Applying the initial conditions we get $D=\operatorname{Id}$ and $g(t)=\exp(tX)$ . ∎

Example 1.2.10.

The map $\mathbb{R}\to\mathrm{SO}(3)=\{g\in\operatorname{GL}_{3}(\mathbb{R})\,|\,gg^{T}% =\operatorname{Id},\,\det g=1\}\subseteq\operatorname{GL}_{3}(\mathbb{R})$ taking $\theta$ to rotation by $\theta$ about a fixed axis is a one-parameter subgroup. Problem 1.2.3 asks you to find its infinitesimal generator.

1.2.3. Exercises

Problem 3. Let $\mathfrak{n}$ be the $\mathbb{C}$ -vector space of strictly upper triangular matrices ( $0$ ’s on the diagonal) and let $N=\{g\in\operatorname{GL}_{n}(\mathbb{C})\,|\,g=\operatorname{Id}+X,X\in% \mathfrak{n}\}$ .

In this problem we will see that the restriction of the exponential to $\mathfrak{n}$ is a diffeomorphism onto $N$ .

(a)

Let $X\in\mathfrak{n}$ . Show that $X^{n}=0$ .
(b)

Show that $\exp(X)\in N$ for $X\in\mathfrak{n}$ .
(c)

Show that, for $g\in N$ , the logarithm $\log(g)=\sum_{k=1}^{\infty}(-1)^{k+1}\tfrac{(g-I)^{k}}{k}$ is in fact a finite sum (and hence converges).
(d)

Show that $\exp|_{\mathfrak{n}}$ and $\log|_{N}$ are inverses of each other. Hint: this boils down to an identity of formal power series, which you can actually deduce from the corresponding fact over $\mathbb{R}$ .

Problem 4. Using the previous question, fill in the gaps of the proof from the notes that

\exp:\mathfrak{gl}_{n,\mathbb{C}}\rightarrow\operatorname{GL}_{n}(\mathbb{C})

is surjective.

(a)

Show all matrices of the form $D$ have a preimage.
(b)

Show all matrices of the form $U$ have a preimage.
(c)

Show all matrices of the form $DU$ have a preimage.
(d)

Is the exponential map $\exp:\mathfrak{sl}_{2,\mathbb{C}}\rightarrow\operatorname{SL}_{2}(\mathbb{C})$ surjective? What about $\exp:\mathfrak{gl}_{2,\mathbb{R}}\rightarrow\operatorname{GL}^{+}_{2}(\mathbb{% R})$ ?

Problem 5. Let ${\bf v}\in\mathbb{R}^{3}$ be a unit vector and let $f:\mathbb{R}\to SO(3)$ be the map with $f(\theta)$ being rotation by $\theta$ about the axis ${\bf v}$ (the angle is measured anticlockwise as you look along the vector from the origin).

Show that $f$ is a one-parameter subgroup and find its infinitesimal generator in terms of ${\bf v}$ .